HDU 1281 棋盘游戏（二分图匹配）

来源：互联网发布：网络规划设计师视频编辑：程序博客网时间：2024/06/10 03:17

题目链接：点击打开链接

思路：经典二分图建图模型，对于每一个格子，按照行标建一列，列标建一列，然后进行匹配即可，然后尝试删除每条边，再进行匹配看看有没有比原匹配小。复杂度显然会超时。。数据水了吧。

细节参见代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<stack>#include<bitset>#include<cstdlib>#include<cmath>#include<set>#include<list>#include<deque>#include<map>#include<queue>#define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))using namespace std;typedef long long ll;typedef long double ld;const ld eps = 1e-9, PI = 3.1415926535897932384626433832795;const int mod = 1000000000 + 7;const int INF = 0x3f3f3f3f;// & 0x7FFFFFFFconst int seed = 131;const ll INF64 = ll(1e18);const int maxn = 100 + 10;int T,n,m,k,tot,kase = 0;int use[maxn],from[maxn];bool mp[maxn][maxn];bool match(int x) {    for(int i = 1; i <= m; i++)    if(!use[i] && mp[x][i]) {        use[i] = true;        if(from[i] == -1 || match(from[i])) {            from[i] = x;            return true;        }    }    return false;}int hungary(int n) {    tot = 0;    memset(from, -1, sizeof(from));    for(int i = 1; i <= n; i++) {        memset(use, 0, sizeof(use));        if(match(i)) ++tot;    }    return tot;}struct node {    int r, c;}a[maxn*maxn];int main() {    while(~scanf("%d%d%d",&n,&m,&k)) {        memset(mp, false, sizeof(mp));        for(int i = 0; i < k; i++) {            scanf("%d%d",&a[i].r,&a[i].c);            mp[a[i].r][a[i].c] = true;        }        int init = hungary(n), ans = 0;        for(int i = 0; i < k; i++) {            mp[a[i].r][a[i].c] = false;            int cur = hungary(n);            mp[a[i].r][a[i].c] = true;            if(cur < init) ++ans;        }        printf("Board %d have %d important blanks for %d chessmen.\n",++kase,ans,init);    }    return 0;}

1 0